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Given is a function $f(x)$, whose indefinite integral $F(x)=\intop f(x)dx$ is known. I want to solve for $$ \intop xf(x)dx .$$

I want to apply integration by parts. Thew expression is equivalent to $\int u\,dv$ with $$ u=x, \ \ \ dv=f(x)dx $$ which implies $$ du=dx, \ \ \ v=F(x)dx. $$

Now, when solving I get $$ \intop xf(x)dx = x F(x) - F(x) $$ which is incorrect. Can someone please tell in what sense I am misusing integratino by parts here?

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    I don't think it's possible to solve that integral. – 2017-02-21
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    Another problem is that without a specific form for f or a classification there is no guarantee that you would choose x as u, perhaps f is logarithmic...Then you would choose f as u instead. – 2017-02-21

3 Answers 3

9

The rule says

$$\int uv' dx = uv - \int u'v dx$$

which in your case translates to

$$\int xf(x)dx = x F(x) - \int 1\cdot F(x)dx$$

so you were almost correct, but there is still an integral on the right side of the equation.

2

The integration by parts formula is $$\int u dv = u v - \int v du$$ and you missed the integral on that second term on the right.

2

You could try differentiation under the integral sign:

$$\frac d{dt}\int F(xt)\ dx\bigg|_{t=1}=\int xf(x)\ dx$$

So you'll need to solve $\int F(x)\ dx$ to continue further.